1004 Counting Leaves
题目描述
A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
Input Specification:
Each input file contains one test case. Each case starts with a line containing 0<N<100, the number of nodes in a tree, and M (<N), the number of non-leaf nodes. Then M lines follow, each in the format:
ID K ID[1] ID[2] … ID[K]
where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 01.
The input ends with N being 0. That case must NOT be processed.
Output Specification:
For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.
The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output 0 1 in a line.
Sample Input:
2 1
01 1 02
Sample Output:
0 1
思路：
首先分析一下样例输出，样例有两个节点，有一个非叶子节点01，非叶子节点有1个孩子节点02，这棵树有两层，一层是01，一层是02，第一层叶子节点个数为0，第二层叶子节点个数为1，因此输出0 1
本题是典型的广度优先搜索算法，从根节点开始，层层遍历每一层的节点，储存其信息，最后输出即可。
代码：
首先重要的要定义结构体node：
其中level代表该节点的层数，num表示该节点的孩子个数，child是一个可变数组，用于存储其孩子的节点序号
struct node
{
int level;
int num;//numbers of this node's children
int *child;//the array store its child
};
接下来是全局变量
#define MAXN 100
int N,M;//the number of nodes in a tree,the number of non-leaf nodes.
bool falg=true;
node family[MAXN];
int queue[MAXN],pt=1;
其中falg用于判断是否输入的N为0【保证没有错误情况发生】，queue队列是用于广度优先搜索的，pt是这个队列的长度全局指针
初始化函数：
void init()
{
scanf("%d %d",&N,&M);
if(N==0)
{
falg=false;
return;
}
for(int i=1;i<=N;i++)//将所有节点的level暂设为1，所有节点没有孩子
{
family[i].level=0;
family[i].num=0;
}
for(int i=0;i<M;i++)
{
int ID,K,temp;scanf("%d %d",&ID,&K);
//记录节点ID的孩子的个数K，并new一个数组，记录孩子的编号
family[ID].num=K;
family[ID].child=new int[K];//new its child array
for(int i=0;i<K;i++)
{
scanf("%d",&temp);
family[ID].child[i]=temp;
}
}
return;
}
BFS函数：
void BFS()
{
//首先将根节点放入队列
queue[pt++]=1;
for(int i=1;i<pt;i++)//由于每次放入新的节点pt都会增加，这样就实现了队列的方法，i表示队列的头，pt是尾部
{
for(int j=0;j<family[queue[i]].num;j++)//对于i节点的所有孩子节点
{
int child=family[queue[i]].child[j];//记录i节点的第j个孩子节点序号
queue[pt++]=child;//将其放入队列
family[child].level=family[i].level+1;//同时这个孩子的层数是i节点层数+1
}
}
return;
}
最后计算每层有多少叶子节点
void output()
{
if(!falg)
return;
int max=0;
for(int i=1;i<=N;i++)//首先要计算最大层数
{
max=max>family[i].level?max:family[i].level;
}
//根据最大层的个数构建长度为max+1的数组，用于储存每一层的叶子节点的个数
int *level=new int[max+1];
for(int i=1;i<=N;i++)
{
//遍历所有节点，如果某一节点没有孩子，证明其是叶子节点，将对应层的叶子节点个数+1
if(family[i].num==0)
{
level[family[i].level]++;
}
}
//输出所有层的叶子节点个数
for(int i=0;i<max+1;i++)
{
if(i==0)
printf("%d",level[i]);
else
printf(" %d",level[i]);
}
printf("\n");
return;
}
主函数
#include<stdio.h>
int main()
{
init();
BFS();
output();
return 0;
}
git仓库：Counting Leaves